AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
is large enough, in every 
square that a natural number is written on each one of its cells, one can find a subsquare from the main square such that the sum of the numbers is this subsquare is divisible by 
.
is large enough, then for each coloring of the subsets of the set 
with colors, two non-empty disjoint subsets 
and 
exist such that , and 
are of the same color.
is large enough, among any points of plane we can find 
points such that these points have pairwise distinct distances. Can you prove the assertion for 
where 
is a positive real number instead of ?
by 
and 
such that 
.
be a subset of 
such that if 
, then the residue obtained when we divide 
by 
also belongs to .
.
be a random point on the smaller arc 
of the circumcircle of square 
, and let 
be the intersection point of segments 
and 
. The feet of the tangents from point 
to the circumcircle of the triangle 
are 
and 
, where 
is the center of the square. Prove that points , 
, and lie on a single circle.
a regular hexagon inscribed in a circle 
. Let 
be a point inside and 
its polar reflection wrt . The rays 
meet again at 
. Call 
the polygon formed by the vertices . Similarly construct the polygon 
using instead. and are congruent.
with 
, let its incircle be tangent to sides 
, 
, and at , 
, and 
, respectively. The internal angle bisector of 
intersects lines 
and 
at 
and 
, respectively. Let 
and 
be distinct points on side such that 
. Finally, let 
be the circumcircle of 
. is tangent to the circumcircle of 
. is tangent to the incircle of . be an even integer, and let 
denote the number of (positive) divisors of . does not divide 
.
such that 
for all 
of natural numbers such that for every pair of natural numbers 
and 
, the following inequality holds:![\[
\frac{1}{2} < \frac{\gcd(a_r, a_s)}{\gcd(r, s)} < 2
\]](http://latex.artofproblemsolving.com/1/6/7/167679c1707b957d87311298ea5b72347a9bdc45.png)
with an area of 
, with 
. If 
, Hence find the value of 
.
be the orthocenter of an acute triangle .
,
,
intersects the circumcircle of at point 
,
,
,respectively.
.
is determined by two angles and . and is![\[ \{(B,C):0<B,C<\pi/2,\; B+C > \pi/2\} \subset \mathbb{R}^2 \]](http://latex.artofproblemsolving.com/c/7/1/c71d5d59c8f59a5f06aa86f75a8d8d74aa1abc50.png)
which is a connected region. Therefore, the range of is connected, hence it's an interval.
: the radius of the circumcircle. Then,![\[AH = 2R \cos A,\quad AA' = 2R \sin(\angle ABA') = 2R \sin \left (B + \frac{\pi}{2} - C \right) = 2R \cos (B-C) \]](http://latex.artofproblemsolving.com/5/d/f/5dfba8c2215eb46bcef1387acf396b0c718a22eb.png)
Let 
, 
, 
. Note that 
.
. Note that is a strictly convex function.
Equality holds iff 
, i.e.
is equilateral. is a strictly convex function,
So,
Every value arbitrary close to 
is obtained as 
, so 
.
, 
, and 
to be the lengths of the altitudes from A, B and C respectively.![$$\frac{AH}{AA'} = \frac{AH}{2h_A-AH} = \frac{2h_A}{2h_A-AH}-1 = \frac{4[ABC]}{4[ABC]-a \cdot AH}-1$$](http://latex.artofproblemsolving.com/6/d/b/6db72abb6a63687f7320abac2336467e5567f257.png)
.![$$ \frac{4[ABC]}{4[ABC]-a \cdot AH}+ \frac{4[ABC]}{4[ABC]-b \cdot BH}+ \frac{4[ABC]}{4[ABC]-c \cdot CH}-3$$](http://latex.artofproblemsolving.com/b/8/0/b80a74c831fa3be254969c2def9bc439ac678b9c.png)
By AM-HM and noticing ![$a\cdot AH + b\cdot BH + c\cdot CH = 4[ABC]$](http://latex.artofproblemsolving.com/6/a/e/6aeffe754c737acd4ce01974434dfc63ca86a113.png)
, we get ![$$ \frac{4[ABC]}{4[ABC]-a \cdot AH}+ \frac{4[ABC]}{4[ABC]-b \cdot BH}+ \frac{4[ABC]}{4[ABC]-c \cdot CH}-3 \geq \frac{9}{2} -3 = \frac{3}{2}$$](http://latex.artofproblemsolving.com/d/1/d/d1dcec89d7b9e63a13668f8cb1dc1ec7f82a3a21.png)
With equality when is equilateral.
and 
.
such that:
